题目：给定元素a,b,a,b,c,c,d，求解出所有的排列。
思路：首先这道题的算法是一个比较经典的算法，它并不是使用分治法思想，也不是使用动态规范的思想，当然更不是贪心的思想。他采用的是穷举法（当然你也可以称之为强力法）！！！之所以强调这个特点，是想提醒自己，对于涉及到数学的概念的题目，往往需要采用强力法，即便使用计算机的思想也不能从根本上降低算法复杂度，除非问题本身存在一个优化的数学解法，比如最大公因子求法，这就是使用辗转相除的方式。
对于本题而言，我们有个数学概念可以改良算法，但是，这个算法并没有本质上降低算法的复杂度，所以，我们还是采用传统的计算机的方法求解全排列。我门借用动态规划思想的备忘录方法进行求解。说到备忘录我想大家就会想到递归，我们就是采用这种递归求解的思想。
代码如下：
#include
#include
#include
using namespace std;
void AllPermutation(vector & vecElementSet, int s, int e, vector & vecAllPermutation)
{
if (s == e) {
vecAllPermutation.push_back(string(vecElementSet.begin(), vecElementSet.end()));
return ;
}
AllPermutation(vecElementSet, s+1, e, vecAllPermutation);
for (int i=s+1; i AllPermutation(vector vecElementSet)
{
string strPermutation;
vector vecAllPermutation;
sort(vecElementSet.begin(), vecElementSet.end());
AllPermutation(vecElementSet, 0, vecElementSet.size(), vecAllPermutation);
return vecAllPermutation;
}
int main(int argc, char ** argv)
{
vector vecElementSet = {
'a', 'b', 'a', 'b', 'c', 'c', 'd'
};
auto vecAllPermutation = AllPermutation(vecElementSet);
for (auto itr=vecAllPermutation.begin(); itr!=vecAllPermutation.end(); ++itr) {
cout << *itr << endl;
}
cout << vecAllPermutation.size() << endl;
// sort(vecAllPermutation.begin(), vecAllPermutation.end());
// vecAllPermutation.erase(unique(vecAllPermutation.begin(), vecAllPermutation.end()), vecAllPermutation.end());
// cout << vecAllPermutation.size() << endl;
//
return 0;
}
该算法对于交换部分是重点，这部分保证了原来重复的元素依然保持连续重复，只有保证了这点，算法才能有效的使用continue的条件。去重复的根本是去掉所有的重复交换！！！（这是参考字典序的算法改进的，该算法的效率不是很高，但是没有更好的办法了）
其实，挺感慨的，学了这么多年的计算机，到头来还是要学习一些数学基本功，但是这些基本功早就还给老师，真的很惭愧，现在连一个高中的排列组合题都不会做了！！！真心不知道自己能在这条路上走多远。
输出：
aabbccd
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